Question

The number of all the possible integral values of $n>\,2$ such that $\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}$ is

• A. 5
• B. 4
• C. 3
• D. infinity

Answer 1

Answer: (C)

Given,
$\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}\,\,\,\,\,\,\,\dots(i)$
Squaring both sides
$\left(\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}\right)^{2}=\frac{n}{4}$
$\sin ^{2} \frac{\pi}{2 n}+\cos ^{2} \frac{\pi}{2 n}+2 \sin \frac{\pi}{2 n} \cos \frac{\pi}{2 n}=\frac{n}{4}$
$\Rightarrow \, 1+\sin \frac{\pi}{n}=\frac{n}{4}$
$ \Rightarrow \,\sin \frac{\pi}{n}=\frac{n}{4}-1$
$ \Rightarrow \, \sin \frac{\pi}{n}=\frac{n-4}{4}$
As, $\sin \frac{\pi}{n}>\,0, n>\,2$
So, $\frac{n-4}{4}>\,0$
$ \Rightarrow \,n>\,4\,\,\,\,\,\,\,\dots(ii)$
Also, $\sin \left(\frac{\pi}{2 n}\right)+\cos \frac{\pi}{2 n}=\sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)$.
From Eq. (i)
$\sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2} \quad \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2 \sqrt{2}}$
Since, $\sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)<\,1, \forall n>\,2$
We get, $\frac{\sqrt{n}}{2 \sqrt{2}}<\,1$
$ \Rightarrow \,n<\,8\,\,\,\,\,\,\,\dots(iii)$
From Eqs. (ii) and (iii) $4<\,n<\,8$
So, $n=5,6,7$
Hence, number of integral value of $n$ is 3